Lab8 - MATH4573
Ryan Gamble
3/29/2021
Task 2
## [1] 4.6496773 1.2519353 0.6686497 3.9725222 0.1594749 3.7261613 1.0247528
## [8] 0.4328739 1.7592587 1.8096780## [1] 19.45498y = (5+0)/2
sigsq = ((5-0)^2) / 12
ybar = mean(sample)
ssq = sd(sample) ^ 2
sprintf("Distribution Mean: %.4f Distribution Variance: %.4f", y, sigsq)## [1] "Distribution Mean: 2.5000 Distribution Variance: 2.0833"## [1] "Sample Mean: 1.9455 Sample Variance: 2.5647"Conclusion: The sample mean and variance are not equal to the population or distribution mean and variance but are close.
\(\begin{equation}E(T)=nE(Y_{i})=n\mu=10\cdot 2.5=25.0\end{equation}\) \(\begin{equation}V(T)=nV(Y_{i})=10\cdot 2.0833 = 20.833\end{equation}\)
\(\begin{equation}E(\bar{Y})=E(Y)=\mu=2.5\end{equation}\) \(\begin{equation}V(\bar{Y})=\frac{\sigma ^{2}}{n}=\frac{2.0833}{10}=0.20833\mu\end{equation}\)
myclt=function(n,iter){
y=runif(n*iter,0,5) # A
data=matrix(y,nr=n,nc=iter,byrow=TRUE) #B
sm=apply(data,2,sum) #C
hist(sm)
sm
}
w=myclt(n=10,iter=10000) #D
## [1] 24.96373## [1] 21.0491myclt_modified=function(n,iter){
y=runif(n*iter,0,5)
data=matrix(y,nr=n,nc=iter,byrow=TRUE)
sm=apply(data,2,mean)
hist(sm)
sm
}
w=myclt_modified(n=10,iter=10000)
## [1] 2.500632Creates iter samples with n elements within the sample from a uniform distribution with a lower limit of 0 and upper limit of 6. These values are stored in one list of size n*iter.
Groups the samples into a matrix where each sample is a column.
Apply the sum function to the columns in the matrix. This is going to return the sum of each sample. It will be size = iter.
Calls the function for a sample population of 10,000 samples of size 10. It stores the output in w.
Task 3
mycltu=function(n,iter,a=0,b=10){
## r-random sample from the uniform
y=runif(n*iter,a,b)
## Place these numbers into a matrix
## The columns will correspond to the iteration and the rows will equal the sample size n
data=matrix(y,nr=n,nc=iter,byrow=TRUE)
## apply the function mean to the columns (2) of the matrix
## these are placed in a vector w
w=apply(data,2,mean)
## We will make a histogram of the values in w
## How high should we make y axis?
## All the values used to make a histogram are placed in param (nothing is plotted yet)
param=hist(w,plot=FALSE)
## Since the histogram will be a density plot we will find the max density
ymax=max(param$density)
## To be on the safe side we will add 10% more to this
ymax=1.1*ymax
## Now we can make the histogram
hist(w,freq=FALSE, ylim=c(0,ymax), main=paste("Histogram of sample mean",
"\n", "sample size= ",n,sep=""),xlab="Sample mean")
## add a density curve made from the sample distribution
lines(density(w),col="Blue",lwd=3) # add a density plot
## Add a theoretical normal curve
curve(dnorm(x,mean=(a+b)/2,sd=(b-a)/(sqrt(12*n))),add=TRUE,col="Red",lty=2,lwd=3) # add a theoretical curve
## Add the density from which the samples were taken
curve(dunif(x,a,b),add=TRUE,lwd=4)
}Apply function uses the parameter 2 to indicate we want to apply the function to the columns.
When n=20 and iter=100,0000. The size of w is 100,000.
Conclusion: As sample size increases, or sampling distribution gets closer to normal.
Task 4
mycltb=function(n,iter,p=0.5,...){
## r-random sample from the Binomial
y=rbinom(n*iter,size=n,prob=p)
## Place these numbers into a matrix
## The columns will correspond to the iteration and the rows will equal the sample size n
data=matrix(y,nr=n,nc=iter,byrow=TRUE)
## apply the function mean to the columns (2) of the matrix
## these are placed in a vector w
w=apply(data,2,mean)
## We will make a histogram of the values in w
## How high should we make y axis?
## All the values used to make a histogram are placed in param (nothing is plotted yet)
param=hist(w,plot=FALSE)
## Since the histogram will be a density plot we will find the max density
ymax=max(param$density)
## To be on the safe side we will add 10% more to this
ymax=1.1*ymax
## Now we can make the histogram
## freq=FALSE means take a density
hist(w,freq=FALSE, ylim=c(0,ymax),
main=paste("Histogram of sample mean","\n", "sample size= ",n,sep=""),
xlab="Sample mean",...)
## add a density curve made from the sample distribution
#lines(density(w),col="Blue",lwd=3) # add a density plot
## Add a theoretical normal curve
curve(dnorm(x,mean=n*p,sd=sqrt(p*(1-p))),add=TRUE,col="Red",lty=2,lwd=3)
}
w=mycltb(n=3,iter=10000, p=0.3)
Conclusion: Again, as the sample size increases, we get a sample distribution closer to normal.
Task 5
mycltp=function(n,iter,lambda=10,...){
## r-random sample from the Poisson
y=rpois(n*iter,lambda=lambda)
## Place these numbers into a matrix
## The columns will correspond to the iteration and the rows will equal the sample size n
data=matrix(y,nr=n,nc=iter,byrow=TRUE)
## apply the function mean to the columns (2) of the matrix
## these are placed in a vector w
w=apply(data,2,mean)
## We will make a histogram of the values in w
## How high should we make y axis?
## All the values used to make a histogram are placed in param (nothing is plotted yet)
param=hist(w,plot=FALSE)
## Since the histogram will be a density plot we will find the max density
ymax=max(param$density)
## To be on the safe side we will add 10% more to this
ymax=1.1*ymax
## Make a suitable layout for graphing
layout(matrix(c(1,1,2,3),nr=2,nc=2, byrow=TRUE))
## Now we can make the histogram
hist(w,freq=FALSE, ylim=c(0,ymax), col=rainbow(max(w)),
main=paste("Histogram of sample mean","\n", "sample size= ",n," iter=",iter," lambda=",lambda,sep=""),
xlab="Sample mean",...)
## add a density curve made from the sample distribution
#lines(density(w),col="Blue",lwd=3) # add a density plot
## Add a theoretical normal curve
curve(dnorm(x,mean=lambda,sd=sqrt(lambda/n)),add=TRUE,col="Red",lty=2,lwd=3) # add a theoretical curve
# Now make a new plot
# Since y is discrete we should use a barplot
barplot(table(y)/(n*iter),col=rainbow(max(y)), main="Barplot of sampled y", ylab ="Rel. Freq",xlab="y" )
x=0:max(y)
plot(x,dpois(x,lambda=lambda),type="h",lwd=5,col=rainbow(max(y)),
main="Probability function for Poisson", ylab="Probability",xlab="y")
}
mycltp(n=2,iter=10000, lambda=4)
